## circumcevian reflections

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• Dear friends, First of all I wish you a very happy and fruitful 2005. To start the new year let me offer a little theorem: If A B C is a circumcevian
Message 1 of 8 , Jan 5, 2005
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Dear friends,

First of all I wish you a very happy and fruitful 2005.

To start the new year let me offer a little theorem:

If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of
A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")
and (AB"C") are concurrent in one point.

Kind regards,
Floor.
• Dear Floor and friends, may I wish all the best to all of you for this new year. ... your point is the isogonal conjugate of the infinite point of the
Message 2 of 8 , Jan 5, 2005
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Dear Floor and friends,

may I wish all the best to all of you for this new year.

> [FvL]To start the new year let me offer a little theorem:
>
> If A'B'C' is a circumcevian triangle [of point P], and A"B"C" are the
> reflections of
> A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
> (A"BC")
> and (AB"C") are concurrent in one point.

your point is the isogonal conjugate of the infinite point of the
direction which is perpendicular to HP.

notice that ABC and A"B"C" are perspective iff P lies on the Jerabek
hyperbola or at infinity.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Antreas, ... No, you mixed up with (A BC), (AB C) and (ABC ) are concurrent in one point (H). Kind regards, Floor.
Message 3 of 8 , Jan 5, 2005
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Dear Antreas,

> [FvL]:
> >If A'B'C' is a circumcevian triangle, and A"B"C" are the reflections of
> >A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C), (A"BC")
> >and (AB"C") are concurrent in one point.
>
> Isn't it equivalent to this:
>
> The circumcircle of ABC and its reflections on the sidelines
> of ABC are concurrent ?

No, you mixed up with (A"BC), (AB"C) and (ABC") are concurrent in one point
(H).

Kind regards,
Floor.
• Dear Bernard, ... [BG] ... Thanks, Bernard! The locus of the perspector for P on the Jerabek hyperbola is a quartic, the isogonal conjugate of the conic SUM
Message 4 of 8 , Jan 6, 2005
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Dear Bernard,

> > [FvL]To start the new year let me offer a little theorem:
> >
> > If A'B'C' is a circumcevian triangle [of point P], and A"B"C" are the
> > reflections of
> > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
> > (A"BC")
> > and (AB"C") are concurrent in one point.

[BG]
> your point is the isogonal conjugate of the infinite point of the
> direction which is perpendicular to HP.
>
> notice that ABC and A"B"C" are perspective iff P lies on the Jerabek
> hyperbola or at infinity.

Thanks, Bernard!

The locus of the perspector for P on the Jerabek hyperbola is a quartic, the
isogonal conjugate of the conic

SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
cyclic

Kind regards,
Floor.
• Dear Floor, ... it seems that your conic decomposes into two secant lines at X110, passing through X112 and X351. please do check this Best regards Bernard
Message 5 of 8 , Jan 6, 2005
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Dear Floor,

> [FvL] The locus of the perspector for P on the Jerabek hyperbola is a
> quartic, the
> isogonal conjugate of the conic
>
> SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
> cyclic

it seems that your conic decomposes into two secant lines at X110,
passing through X112 and X351.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, ... [BG] ... This is caused by a typo. The conic should be SUM b^4c^4SA(b^2-c^2)^4x^2 - a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0 cyclic So with - in
Message 6 of 8 , Jan 6, 2005
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Dear Bernard,

> > [FvL] The locus of the perspector for P on the Jerabek hyperbola is a
> > quartic, the
> > isogonal conjugate of the conic
> >
> > SUM b^4c^4SA(b^2-c^2)^4x^2 + a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
> > cyclic

[BG]
> it seems that your conic decomposes into two secant lines at X110,
> passing through X112 and X351.
>

This is caused by a typo. The conic should be

SUM b^4c^4SA(b^2-c^2)^4x^2 - a^6b^2c^2(c^2-a^2)^2(a^2-b^2)^2yz = 0
cyclic

So with - in stead of +. My apologies.

Kind regards,
Floor.
• Dear Floor ... Happy New Year to you and to each Hyacinthist. ... reflections of ... (A BC ) ... Note that if A ,B ,C lie on the circumcircle, the circles
Message 7 of 8 , Jan 6, 2005
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Dear Floor
> First of all I wish you a very happy and fruitful 2005.

Happy New Year to you and to each Hyacinthist.

> To start the new year let me offer a little theorem:
> If A'B'C' is a circumcevian triangle, and A"B"C" are the
reflections of
> A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
(A"BC")
> and (AB"C") are concurrent in one point.

Note that if A',B',C' lie on the circumcircle, the circles A"B"C,
B"C"A, C"A"B have a common point.
I suspect that this common point lies on the circumcircle if and
only if the lines AA', BB', CC' concur.
Any idea?
Friendly. Jean-Pierre
• Dear Jean-Pierre, [FvL] ... [JPE] ... If we have A and B , we can construct A , B , and then (A B C), and after that the common point P of (A B C) and the
Message 8 of 8 , Jan 7, 2005
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Dear Jean-Pierre,

[FvL]
> > To start the new year let me offer a little theorem:
> > If A'B'C' is a circumcevian triangle, and A"B"C" are the
> reflections of
> > A'B'C' through the sides of ABC, then the circles (ABC), (A"B"C),
> (A"BC")
> > and (AB"C") are concurrent in one point.

[JPE]
> Note that if A',B',C' lie on the circumcircle, the circles A"B"C,
> B"C"A, C"A"B have a common point.
> I suspect that this common point lies on the circumcircle if and
> only if the lines AA', BB', CC' concur.
> Any idea?

If we have A' and B', we can construct A", B", and then (A"B"C), and after
that the common point P of (A"B"C) and the circumcircle. That means we have
the circles (AB"P)=(AB"C") and (A"BP)=(A"BC") as well. The point C" is thus
fixed as the second intersection of these circles (apart from P). So only
one point C", and consequently one point C' satisfies the condition. This
must be the circum-cevian trace of the intersection of AA' and BB'.

Kind regards,
Floor.
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